# Minimize swaps to rearrange array such that parity of index and corresponding element is same

• Last Updated : 10 Mar, 2022

Given an array A[], the task to find minimum swapping operations required to modify the given array A[] such that for every index in the array, parity(i) = parity(A[i]) where parity(x) = x % 2. If it’s impossible to obtain such an arrangement, then print -1.

Examples:

Input: A[] = { 2, 4, 3, 1, 5, 6 }
Output:
Explanation:
Swapping (4, 3) and (5, 6) modifies the array to [2, 3, 4, 1, 6, 5] such that the parity of i and A[i] is same for all indices.

Input: A[] = {1, 2, 5, 7}
Output: -1
Explanation:
The given array cannot be rearranged as per required condition.

Approach:
To solve the problem mentioned above an optimal approach is to choose such an index where parity(i) and parity(A[i]) aren’t the same.

• Initialize two variables needodd and needeven to 0 which will store the parity of each element. Check the parity of index if it’s odd then increase needodd value by 1 otherwise increase needeven.
• If needodd and needeven aren’t same, then required arrangement is not possible.
• Otherwise, the final result is obtained by needodd variable, as it is the number of operations that are required. This is because, at any moment, we choose an odd element whose parity is not the same with parity of their index and similarly choose an even element and swap them.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to minimize``// swaps required to rearrange``// array such that parity of index and``// corresponding element is same``#include ``using` `namespace` `std;` `// Function to return the``// parity of number``int` `parity(``int` `x)``{``    ``return` `x % 2;``    ` `}` `// Function to return minimum``// number of operations required``int` `solve(``int` `a[], ``int` `size)``{``    ` `    ``// Initialize needodd and``    ``// needeven value by 0``    ``int` `needeven = 0;``    ``int` `needodd = 0;``    ` `    ``for``(``int` `i = 0; i < size; i++)``    ``{``        ``if``(parity(i) != parity(a[i]))``        ``{``            ` `            ``// Check if parity(i) is odd``            ``if``(parity(i) % 2)``            ``{``                ` `                ``// increase needodd``                ``// as we need odd no``                ``// at that position.``                ``needodd++;``            ``}``            ``else``            ``{` `                ``// increase needeven``                ``// as we need even``                ``// number at that position``                ``needeven++;``            ``}``        ``}``    ``}``    ` `    ``// If needeven and needodd are unequal``    ``if``(needeven != needodd)``        ``return` `-1;``    ``else``        ``return` `needodd;``}` `// Driver Code``int` `main()``{``    ``int` `a[] = { 2, 4, 3, 1, 5, 6};``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);` `    ``// Function call``    ``cout << solve(a, n) << endl;` `    ``return` `0;``}` `// This code is contributed by venky07`

## Java

 `// Java implementation to minimize ``// swaps required to rearrange ``// array such that parity of index and ``// corresponding element is same``import` `java.util.*;` `class` `GFG{``    ` `// Function to return the``// parity of number``static` `int` `parity(``int` `x)``{``    ``return` `x % ``2``;``}` `// Function to return minimum``// number of operations required``static` `int` `solve(``int` `a[], ``int` `size)``{``    ` `    ``// Initialize needodd and``    ``// needeven value by 0``    ``int` `needeven = ``0``;``    ``int` `needodd = ``0``;``    ` `    ``for``(``int` `i = ``0``; i < size; i++)``    ``{``        ``if``(parity(i) != parity(a[i]))``        ``{``            ` `            ``// Check if parity(i) is odd``            ``if``(parity(i) % ``2` `== ``1``)``            ``{``                ` `                ``// Increase needodd``                ``// as we need odd no``                ``// at that position.``                ``needodd++;``            ``}``            ``else``            ``{``                ` `                ``// Increase needeven``                ``// as we need even``                ``// number at that position``                ``needeven++;``            ``}``        ``}``    ``}``    ` `    ``// If needeven and needodd are unequal``    ``if``(needeven != needodd)``        ``return` `-``1``;``    ``else``        ``return` `needodd;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``int` `a[] = { ``2``, ``4``, ``3``, ``1``, ``5``, ``6``};``    ``int` `n = a.length;``    ` `    ``// Function call``    ``System.out.println(solve(a, n));``}``}` `// This code is contributed by offbeat`

## Python3

 `# Python3 implementation to minimize``# swaps required to rearrange``# array such that parity of index and``# corresponding element is same` `# Function to return the``# parity of number``def` `parity(x):``    ``return` `x ``%` `2` `# Function to return minimum``# number of operations required` `def` `solve(a, size):``    ` `    ``# Initialize needodd and``    ``# needeven value by 0``    ``needeven ``=` `0``    ``needodd ``=` `0``    ``for` `i ``in` `range``(size):``                ` `        ``if` `parity(i)!``=` `parity(a[i]):``            ` `            ``# Check if parity(i) is odd``            ``if` `parity(i) ``%` `2``:``                ` `                ``# increase needodd``                ``# as we need odd no``                ``# at that position.``                ``needodd``+``=` `1``            ``else``:``                ``# increase needeven``                ``# as we need even``                ``# number at that position``                ``needeven``+``=` `1``    ` `    ``# If needeven and needodd are unequal``    ``if` `needodd !``=` `needeven:``        ``return` `-``1``        ` `    ``return` `needodd``    ` `# Driver code    ``if` `__name__ ``=``=``"__main__"``:``    ` `    ``a ``=` `[``2``, ``4``, ``3``, ``1``, ``5``, ``6``]``    ``n ``=` `len``(a)``    ``print``(solve(a, n))`

## C#

 `// C# implementation to minimize ``// swaps required to rearrange ``// array such that parity of index and ``// corresponding element is same``using` `System;``class` `GFG{``    ` `// Function to return the``// parity of number``static` `int` `parity(``int` `x)``{``  ``return` `x % 2;``}` `// Function to return minimum``// number of operations required``static` `int` `solve(``int``[] a, ``int` `size)``{    ``  ``// Initialize needodd and``  ``// needeven value by 0``  ``int` `needeven = 0;``  ``int` `needodd = 0;` `  ``for``(``int` `i = 0; i < size; i++)``  ``{``    ``if``(parity(i) != parity(a[i]))``    ``{``      ``// Check if parity(i) is odd``      ``if``(parity(i) % 2 == 1)``      ``{``        ``// Increase needodd``        ``// as we need odd no``        ``// at that position.``        ``needodd++;``      ``}``      ``else``      ``{``        ``// Increase needeven``        ``// as we need even``        ``// number at that position``        ``needeven++;``      ``}``    ``}``  ``}` `  ``// If needeven and needodd are unequal``  ``if``(needeven != needodd)``    ``return` `-1;``  ``else``    ``return` `needodd;``}` `// Driver Code``public` `static` `void` `Main ()``{``  ``int``[] a = {2, 4, 3, 1, 5, 6};``  ``int` `n = a.Length;` `  ``// Function call``  ``Console.Write(solve(a, n));``}``}` `// This code is contributed by Chitranayal`

## Javascript

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Output:

`2`

Time Complexity: O(n), where n is the size of the given array
Auxiliary Space: O(1)

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